A chemist pours 200 grams of a liquid (density = 2.5 gm/mL) into a cube beaker with 5 cm edges and a graduated cylinder with a base radius of 5 cm and a height of 5 cm.
What is fraction of the liquid that needs to be put in the graduated cylinder such that the liquid height is the same in the beaker
and graduated cylinder?

The area of the graduated cylinder (g) floor = pi*5*5 = 25*pi.
Th area of the beaker (b) floor is 5*5 = 25.
(Area)g/(Area)b = 25*pi/25 = pi
Since the area of the graduated cylinder is larger than that of the beaker, the volume poured into the graduated cylinder must be pi times volume of that poured into the beaker.
Let x = volume poured into the beaker, then 80-x = volume poured into the graduated cylinder, and
pi*x = 80-x
Solve for x, then x/80 = fraction.
I get x = about 19 and 80-x = about 61.
Check and solve for h for grad cyl = pi*r^2*h and
solve for h for beaker from V= 25*h; i.e.,
h = Vg/pi*25 = about 0.8
h = Vb/25 = about 0.8
You can do it more accurately. Check my work. I'm not a math man. Perhaps Reiny or Mathmate will check it for us.